For what value of k the following system of equations has a unique solution `2x+3y-5=0, kx-6y-8=0?`

Options

A. `k=-4`
B. `k!=-4`
C. `k!=4`
D. `k=4`

Answer

To find the value of \( k \) for which the system of equations has a unique solution, we need to analyze the given equations:

1. \( 2x + 3y - 5 = 0 \) (Equation 1)
2. \( kx - 6y - 8 = 0 \) (Equation 2)

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = 2 \)
- \( b_1 = 3 \)
- For Equation 2:
- \( a_2 = k \)
- \( b_2 = -6 \)

### Step 2: Condition for unique solution
For a system of linear equations to have a unique solution, the following condition must hold:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Substituting the coefficients we identified:
\[ \frac{2}{k} \neq \frac{3}{-6} \]

### Step 3: Simplify the right side
Simplifying the right side:
\[ \frac{3}{-6} = -\frac{1}{2} \]
So, the condition becomes:
\[ \frac{2}{k} \neq -\frac{1}{2} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2 \cdot (-2) \neq k \cdot 1 \]
This simplifies to:
\[ -4 \neq k \]

### Step 5: Conclusion
Thus, for the system of equations to have a unique solution, \( k \) must not equal \(-4\). Therefore, the final answer is:
\[ k \neq -4 \]